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Genrish500 [490]
3 years ago
6

Can someone help me solve this math problem?

Mathematics
1 answer:
RoseWind [281]3 years ago
4 0
The correct answer is f(6) = 14.

f(x) is essentially the y-value, and the question is asking you:

if the value of x is 6, its corresponding y-value (or f(x)) is 14.
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Puteti sa ma ajutati va rog? Vreau solutii cu explicatie elaborata, nu doar raspunsul. Multumesc.
almond37 [142]
I don’t speak Spanish but if you put the English trans in the comments I’d love to answer
6 0
3 years ago
What are th zeros of f(x)=x(x-8)
Andrews [41]
Hi there!

Let's solve this problem step by step!
The zeros of a function are the input values that produce an outcome of 0. To find this input values we must set up and solve an equation.

f(x) = x(x - 8)
Set up the equation

f(x) = 0 thus x(x - 8) = 0

Now use the rule AB = 0 gives A = 0 or B = 0. This rule basically means that if we have a product that equals 0, at least one of the factors must be 0.

Hence, we get the following:
x = 0 \: \: or \: \: x - 8 = 0

Finally add 8 to both sides in the right equation.
x = 0 \: \: or \: \: x = 8

The zeros of the function are 0 and 8.
~ Hope this helps you!
4 0
3 years ago
17. Does this graph below represent a linear or nonlinear function? ^^
TiliK225 [7]

Answer:linear

Step-by-step explanation:

The numbers are going up at the same rate and it’s a straight line

5 0
3 years ago
Read 2 more answers
Butch, Harry, and Nicky are playing a game. Harry has half as many points as
ryzh [129]

Answer:

9

Step-by-step explanation:

H=1/2 * B

N=H-5= 1/2 * B-5

B+H+N= B+(1/2 *B) + (1/2 *B-5)=31

2B-5=31

B=18

H=1/2 *B=9

4 0
4 years ago
The center of a cricle ids (h,7) and the radius is 10. The circle passes through (3,-1). Find all possible values of h.
Nostrana [21]
center\ of\ a\ circle:(a;\ b)\\\\radius:r\\\\(x-a)^2+(y-b)^2=r^2\\\\========================\\\\(h;\ 7);\ r=10\\\\(x-h)^2+(y-7)^2=10^2\\\\========================

The\ circle\ passes\ throught\ (3;-1).\\\\Substitute\ x=3\ and\ y=-1:\\\\(3-h)^2+(-1-7)^2=100\\\\(3-h)^2+(-8)^2=100\\\\(3-h)^2+64=100\ \ \ /-64\\\\(3-h)^2=36\iff3-h=\pm\sqrt{36}\\\\3-h=-6\ or\ 3-h=6\\-h=-6-3\ or\ -h=6-3\\-h=-9\ or\ -h=3\\h=9\ or\ h=-3\\\\Answer:h=9\ or\ h=-3
6 0
3 years ago
Read 2 more answers
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